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    Argument of complex number - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • 25 Questions around this concept.

    Solve by difficulty

    If general argument of a complex no is $2n\pi +\frac{4\pi }{3};$ what is the principle argument

    The argument of the complex no. $\frac{2+3 i}{3+i+(1+2 i)^2}$ is

    Principal value of amplitude of $(1+i)$ is:

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    Principal value of the argument of $\cos 1200^{o}+i \sin 1200^{o}$ is:  

    The point represented by the complex number $2 -i$ is rotated about origin through an angle $\pi/2$ in the clockwise direction, the new position of point is:

    Let $O$ be the origin, the point $A$ be $z_1=\sqrt{3}+2 \sqrt{2} i$, the point $B\left(z_2\right)$ be such that $\sqrt{3}\left|z_2\right|=\left|z_1\right|$ and $\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6}$. Then

    Concepts Covered - 1

    Argument of complex number

    If a complex number z = x + iy is represented by a point P in the Argand plane and OP forms some angle with a positive x-axis, let's denote it with ?, then ? is called the argument of z.

    $\begin{aligned} & \tan \theta=\frac{\mathrm{PM}}{\mathrm{OM}} \\ & \tan \theta=\frac{\mathrm{y}}{\mathrm{x}}=\frac{\operatorname{Im}(\mathrm{z})}{\operatorname{Re}(\mathrm{z})} \Rightarrow \theta=\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}} \\ & \arg (\mathrm{z})=\theta=\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}\end{aligned}$

    If ? lies between -? < ? ≤ ?, then ? is called the principal argument. The value of the argument differs depending on which quadrant point (x,y) lies.

    If it lies in 1st quadrant then it is ? (acute angle)


     

     

    If the point lies in 2nd quadrant, then  $\arg (z)=\theta=\pi-\tan ^{-1} \frac{y}{|x|}$ 

    So it will be an obtuse +ve angle

    If the point lies in lies in 3rd quadrant then  $\arg (z)=\theta=-\pi+\tan ^{-1} \frac{y}{x}$

    It will be an obtuse -ve angle

     

    If the point lies in 4th quadrant then  $\arg (z)=\theta=-\tan ^{-1} \frac{|y|}{x}$

    It will be a -ve acute angle

    Note:

    If $\arg (\mathrm{z})=\frac{\pi}{2}$ or $-\frac{\pi}{2}, \mathrm{z}$ is purely imaginary.
    If $\arg (\mathrm{z})=0$ or $\pi, \mathrm{z}$ is purely real.

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    Argument of complex number

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